Question

In the reaction: $ C{{F}_{3}}-CH=C{{H}_{2}}+HCl\to $
Which one of the following is formed as the major product?
(A) $ C{{F}_{3}}C{{H}_{2}}C{{H}_{2}}Cl $
(B) $ \,C{{F}_{3}}CHClC{{H}_{3}} $
(C) $ ClC{{F}_{2}}-CHF-C{{H}_{3}} $
(D) $ CH{{F}_{2}}CHFC{{H}_{2}}Cl $

Answer 1

Hint :We know that hydrogen halide gives an additional reaction to alkene. Alkene behaves as nucleophilic so, gets protonated by hydrogen halide. Then at carbocation halide attacks. So, alkyl halide forms as a product. The formation of the product is decided based on the Markovnikov rule.

Complete Step By Step Answer:
As we know that in organic chemistry, hydrocarbons are an important topic. The hydrocarbons are majorly classified as three groups. There are alkane, alkene and alkyne. The alkane means carbon-carbon single bond. The alkene has a carbon-carbon double bond.
According to this halogen that means the most electronegative atom in the reagent going to addition in the carbon having the least hydrogen atom in alkene or alkyne molecule. The hydrogen in the reagent is going to attack the carbon having more number of hydrogen atoms in the alkene or alkyne. The alkyne means carbon-carbon having triple bonds in the molecule. The reaction of hydrogen chloride with alkene is an example of an electrophilic addition reaction. In the first step, the electrophilic addition takes place. In the second step, the nucleophilic addition takes place.
Hydrogen chloride gives an additional reaction on propene. During the reaction, first, alkene attacks on the hydrogen of hydrogen chloride, so a carbocation forms. On this carbocation, chloride gets attached. According to the Markovnikov rule, the negative part of the attacking group gets attached to the position where less number of hydrogens is present. Thus, the reaction is given by:
 $ C{{F}_{3}}-CH=C{{H}_{2}}+HCl\to C{{F}_{3}}C{{H}_{2}}C{{H}_{2}}Cl $
Therefore, the correct answer is option A.

Note :
Note that Markovnikov rule is based upon the observation of stability of the carbocation which formed during the reaction. In the case of unsymmetrical alkene two possibilities are there for the formation of a carbocation. In secondary carbocation, the more electron-donating groups are present which stabilizes the positive charge of the carbocation. So secondary carbocation is more stable than primary, so the reaction goes through the formation of a secondary carbocation.